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Let $A=\begin{pmatrix} 6&3&-8\\ 0&-2&0\\ 1&0&-3\\ \end{pmatrix}$. I found that the two distinct eigenvalues are $\lambda=5,-2$. Now, I am asked to find a basis for each eigenspace corresponding to each eigenvalue. I found the basis for the eigenspace corresponding to $\lambda=5$ to be $(8,0,1)$. For the eigenspace corresponding to $\lambda=-2$, I get a basis different than the key; below is my work.

We say $\vec{x}$ is an eigenvector of a given matrix A if and only if A$\vec{x}=\lambda\vec{x}$ for some eigenvalue $\lambda$. Thus, we solve for $\vec{x}$. $$(A-\lambda I)\vec{x}=\vec{0}$$ $$\begin{pmatrix} 6&3&-8\\ 0&-2&0\\ 1&0&-3\end{pmatrix}\vec{x}-\begin{pmatrix} -2&0&0\\ 0&-2&0\\ 0&0&-2\\ \end{pmatrix}\vec{x}=\vec{0}$$ $$\begin{pmatrix} 8&3&-8\\ 0&0&0\\ 1&0&-1\\ \end{pmatrix}\vec{x}=\vec{0}$$ This equates to to solving the linear system: $$\begin{pmatrix} 8&3&-8&0\\ 0&0&0&0\\ 1&0&-1&0\\ \end{pmatrix}$$ Interchanging rows $1$ and $3$: $$\begin{pmatrix} 1&0&-1&0\\ 0&0&0&0\\ 8&3&-8&0\\ \end{pmatrix}$$ Eliminating below $1$ in first column: $$\begin{pmatrix} 1&0&-1&0\\ 0&0&0&0\\ 0&3&0&0\\ \end{pmatrix}$$ Interchanging rows $2$ and $3$ and rescaling: $$\begin{pmatrix} 1&0&-1&0\\ 0&1&0&0\\ 0&0&0&0\\ \end{pmatrix}$$ Thus I would conclude my basis for the eigenspace corresponding to $\lambda=-2$ is $(1,0,0)$. However, the book insists the basis is $(1,0,1)$. Are there any errors in my work? Thanks in advance!

Edit: I now see my error. The basis is indeed $(1,0,1)$.

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    $\begingroup$ the step for $\begin{pmatrix} 1&0&-1&0\\ 0&0&0&0\\ 8&3&-8&0\\ \end{pmatrix}$ to $\begin{pmatrix} 1&0&-1&0\\ 0&0&8&0\\ 0&3&0&0\\ \end{pmatrix}$ is not clear and seems to be wrong. $\endgroup$ – gimusi Apr 5 '18 at 14:10
  • $\begingroup$ I added -8R1 to R3. $\endgroup$ – coreyman317 Apr 5 '18 at 14:12
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    $\begingroup$ and what is the 8 in the second row? $\endgroup$ – gimusi Apr 5 '18 at 14:13
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    $\begingroup$ and in the final step what is the 1 in the right up entry? $\endgroup$ – gimusi Apr 5 '18 at 14:14
  • $\begingroup$ Ah, thank you. All-nighters make calculations a nightmare. Thanks for the quick help! $\endgroup$ – coreyman317 Apr 5 '18 at 14:14
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From here,

$$\begin{pmatrix} 8&3&-8\\ 0&0&0\\ 1&0&-1\\ \end{pmatrix}\vec{x}=\vec{0}$$

we see that $\vec{x}=(1,0,1)$ is the correct solution.

Indeed from here

$$\begin{pmatrix} 8&3&-8&0\\ 0&0&0&0\\ 1&0&-1&0\\ \end{pmatrix}\stackrel{R_1-8R_3}\to \begin{pmatrix} 0&3&0&0\\ 0&0&0&0\\ 1&0&-1&0\\ \end{pmatrix} \implies x_2=0 \quad x_1=x_3$$

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  • $\begingroup$ Where did I go wrong in my calculation? $\endgroup$ – coreyman317 Apr 5 '18 at 14:09
  • $\begingroup$ @coreyman317 I've just pointed out in the comment under the OP, second step is not clear and seems uncorrect. $\endgroup$ – gimusi Apr 5 '18 at 14:11

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